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PHYS 1444 – Section 501 Lecture #4

PHYS 1444 – Section 501 Lecture #4. Monday, Jan. 30, 2006 Dr. Jae hoon Yu. Gauss’ Law Electric Flux Generalization of Electric Flux How are Gauss’ Law and Coulomb’s Law Related? Electric Potential Energy Electric Potential. Announcements.

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PHYS 1444 – Section 501 Lecture #4

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  1. PHYS 1444 – Section 501Lecture #4 Monday, Jan. 30, 2006 Dr. Jaehoon Yu • Gauss’ Law • Electric Flux • Generalization of Electric Flux • How are Gauss’ Law and Coulomb’s Law Related? • Electric Potential Energy • Electric Potential PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  2. Announcements • Your 3 extra credit points for e-mail subscription is till midnight tonight. • All but two of you have subscribed so far. • I will send out a test message tonight. • Your prompt confirmation reply is appreciated!!! • Please make sure that you do not “reply all” to the list. Be sure to reply back only to me at jaehoonyu@uta.edu. • All of you have registered in the homework system • All of you have submitted HW1!!! • Very impressive!!! • Some of you worked on HW2 already!! • Reading assignments • CH22–3 and 4 PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  3. Gauss’ Law • Gauss’ law states the relationship between electric charge and electric field. • More general and elegant form of Coulomb’s law. • The electric field by the distribution of charges can be obtained using Coulomb’s law by summing (or integrating) over the charge distributions. • Gauss’ law, however, gives an additional insight into the nature of electrostatic field and a more general relationship between the charge and the field PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  4. Electric Flux • Let’s imagine a surface of area A through which a uniform electric field E passes • The electric flux is defined as • FE=EA, if the field is perpendicular to the surface • FE=EAcosq, if the field makes an angle q to the surface • So the electric flux is defined as . • How would you define the electric flux in words? • Total number of field lines passing through the unit area perpendicular to the field. PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  5. Example 22 – 1 • Electric flux. (a) Calculate the electric flux through the rectangle in the figure (a). The rectangle is 10cm by 20cm and the electric field is uniform with magnitude 200N/C. (b) What is the flux in figure if the angle is 30 degrees? The electric flux is So when (a) q=0, we obtain And when (b) q=30 degrees, we obtain PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  6. Generalization of the Electric Flux • Let’s consider a surface of area A that is not a square or flat but in some random shape, and that the field is not uniform. • The surface can be divided up into infinitesimally small areas of DAi that can be considered flat. • And the electric field through this area can be considered uniform since the area is very small. • Then the electric flux through the entire surface can is approximately • In the limit where DAi  0, the discrete summation becomes an integral. open surface enclosed surface PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  7. Generalization of the Electric Flux • We arbitrarily define that the area vector points outward from the enclosed volume. • For the line leaving the volume, q<p/2, so cosq>0. The flux is positive. • For the line coming into the volume, q>p/2, so cosq<0. The flux is negative. • If FE>0, there is a net flux out of the volume. • If FE<0, there is flux into the volume. • In the above figures, each field that enters the volume also leaves the volume, so • The flux is non-zero only if one or more lines start or end inside the surface. PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  8. Generalization of the Electric Flux • The field line starts or ends only on a charge. • Sign of the net flux on the surface A1? • The net outward flux (positive flux) • How about A2? • Net inward flux (negative flux) • What is the flux in the bottom figure? • There should be a net inward flux (negative flux) since the total charge inside the volume is negative. • The flux that crosses an enclosed surface is proportional to the total charge inside the surface.  This is the crux of Gauss’ law. PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  9. Gauss’ Law • The precise relation between flux and the enclosed charge is given by Gauss’ Law • e0 is the permittivity of free space in the Coulomb’s law • A few important points on Gauss’ Law • Freedom to choose!! • The integral is performed over the value of E on a closed surface of our choice in any given situation. • Test of existence of electrical charge!! • The charge Qencl is the net charge enclosed by the arbitrary closed surface of our choice. • Universality of the law! • It does NOT matter where or how much charge is distributed inside the surface. • The charge outside the surface does not contribute to Qencl. Why? • The charge outside the surface might impact field lines but not the total number of lines entering or leaving the surface PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  10. Gauss’ Law q’ q • Let’s consider the case in the above figure. • What are the results of the closed integral of the gaussian surfaces A1 and A2? • For A1 • For A2 PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  11. Coulomb’s Law from Gauss’ Law • Let’s consider a charge Q enclosed inside our imaginary Gaussian surface of sphere of radius r. • Since we can choose any surface enclosing the charge, we choose the simplest possible one!  • The surface is symmetric about the charge. • What does this tell us about the field E? • Must have the same magnitude at any point on the surface • Points radially outward / inward parallel to the surface vector dA. • The Gaussian integral can be written as Solve for E Electric Field of Coulomb’s Law PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  12. Gauss’ Law from Coulomb’s Law • Let’s consider a single static point charge Q surrounded by an imaginary spherical surface. • Coulomb’s law tells us that the electric field at a spherical surface is • Performing a closed integral over the surface, we obtain Gauss’ Law PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

  13. Gauss’ Law from Coulomb’s LawIrregular Surface • Let’s consider the same single static point charge Q surrounded by a symmetric spherical surface A1 and a randomly shaped surface A2. • What is the difference in the number of field lines passing through the two surface due to the charge Q? • None. What does this mean? • The total number of field lines passing through the surface is the same no matter what the shape of the enclosed surface is. • So we can write: • What does this mean? • The flux due to the given enclosed charge is the same no matter what the shape of the surface enclosing it is.  Gauss’ law, , is valid for any surface surrounding a single point charge Q. PHYS 1444-501, Spring 2006 Dr. Jaehoon Yu

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